Question

TÌM Xa)1-6x²-x=0b)12x²+13x+3=0c)x³-11x²+30x=0d)(x²+x+1)(x²+x+2)-12=0e)2x²+x+6=0

Answer 1

a)1-6x²-x =0<=>-(6x²+x-1)=0<=>6x²+x-1=0

<=>(6x²+3x)-(2x+1)=0<=>3x(2x+1)-(2x+1)=0

<=>(3x-1)(2x+1)=0

=>3x-1=0 hoặc 2x+1=0=>x=\(\dfrac13\) hoặc x=-\(\dfrac12\)

Vậy S={\(\dfrac13\);-\(\dfrac12\)}

b)12x²+13x+3=0<=>12x²+9x+4x+3=0<=>(12x²+9x)+(4x+3)=0

<=>3x(4x+3)+(4x+3)=0<=>(3x+1)(4x+3)=0

=>3x+1=0 hoặc 4x+3=0 <=>x=-\(\dfrac13 \) hoặc x=-\(\dfrac34\)

Vậy S={-\(\dfrac13 \);-\(\dfrac34 \)}

c)x³-11x²+30x=0<=>x(x²-11x+30)=0<=>x[(x²-6x)-(5x-30)]=0

<=>x[x(x-6)-5(x-6)]=0<=>x(x-5)(x-6)=0

=>x=0 hoặc x-5=0 hoặc x-6=0=>x=0 hoặc x=5 hoặc x=6

Vậy S={0;5;6}

d)Ta có:(x²+x+1)(x²+x+2)-12=0

Đặt:t=x²+x+1

Khi đó:a(a+1)-12=0<=>a²+a-12=0<=>(a²+4a)-(3a+12)=0

<=>a(a+4)-3(a+4)=0<=>(a-3)(a+4)=0

hay (x²+x-2)(x²+x+5)=0

<=>(x-1)(x+2)(x²+x+5)=0(x²+x-2=(x-1)(x+2))

=>x-1=0 hoặc x+2=0(vì x²+x+5=(x+\(\dfrac12\))²+\(\dfrac{19}{4}\)>0)

=>x=1 hoặc x=-2

Vậy S={1;-2}

e)Ta có:2x²+x+6>x²+x+6=(x+\(\dfrac12\))²+\(\dfrac{23}{4}\)>0

nên PT vô nghiệm

Vậy S=\(\varnothing\)